Chapter 8.7, Problem 20E

To determine

To Prove: The given series.

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-\frac{\pi }{2}\right)}^{2n}}{\left(2n\right)!}}$ .

Calculation:

Consider the given function is,

  $f\left(x\right)=\sin x$

Then,

  $\begin{array}{l}{f}^{\prime }\left(\frac{\pi }{2}\right)=0\\ f^{″}\left(\frac{\pi }{2}\right)=-1\\ f^{‴}\left(\frac{\pi }{2}\right)=0\\ f^4\left(\frac{\pi }{2}\right)=1\end{array}$

Consider the general form of the McLaurin series is,

  $\begin{array}{c}\left[\begin{array}{l}f\left(\frac{\pi }{2}\right)+\frac{f^{\prime }\left(\frac{\pi }{2}\right)\left(x-\left(\frac{\pi }{2}\right)\right)}{1!}\\ +\frac{f^{″}\left(\frac{\pi }{2}\right)}{2!}{\left(x-\left(\frac{\pi }{2}\right)\right)}^2+\frac{f^{‴}\left(\frac{\pi }{2}\right)x^3}{3!}{\left(x-\left(\frac{\pi }{2}\right)\right)}^3+\frac{f^4\left(\frac{\pi }{2}\right)}{4!}{\left(x-\left(\frac{\pi }{2}\right)\right)}^4+\mathrm{...}\end{array}\right]=\left[\begin{array}{l}1+\frac{1}{2!}{\left(x-\frac{\pi }{2}\right)}^2\\ +\frac{1}{4!}{\left(x-\frac{\pi }{2}\right)}^4+\mathrm{...}\end{array}\right]\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-\frac{\pi }{2}\right)}^{2n}}{\left(2n\right)!}}\end{array}$

Since, $f^{n+1}\left(x\right)$ is $\pm \sin x$ or $\pm \cos x$ .

Then,

  $f^{n+1}\left(\frac{\pi }{2}\right)$ is $\pm 1$ or 0

Therefore, $\left|f^{n+1}\left(\frac{\pi }{2}\right)\right|\le 1$ for all the values of x so take $M=1$

Consider the inequality,

  $\left|R_n\left(x\right)\right|\le \frac{M}{\left(n+1\right)!}\left|{\left(x-\frac{\pi }{2}\right)}^{n+1}\right|=\frac{\left|{\left(x-\frac{\pi }{2}\right)}^{n+1}\right|}{\left(n+1\right)!}$

The above equation evaluated to 0 as n tends to infinity

Thus, the value of $\sin x$ is equal to its series for all x and the series is proved.